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I know this is a very common question but there is something I seem to be misunderstanding. Let $S_n$ be a simple random walk in $\Bbb Z^2$ starting at $0$. I want to count the number of paths of length $2n$ that return to the origin. In $\Bbb Z$, this number is $\binom{2n}{n}$. In $\Bbb Z^2$, such a path must make $k$ steps north, $k$ steps south, $n-k$ steps east, $n-k$ steps west. For me, it chooses $k$ steps north among the $2n$ steps, then it chooses $n-k$ east among the $2n-2k$ steps left. So I got $$\sum_{k=0}^n \binom{2n}{k}\binom{2n-2k}{n-k} \text{ paths}$$

However, even after developing it, it is not the expected answer that is $\binom{2n}{n}^2$ paths. Is there something wrong in my argument ? Also, I really want to count it as a sum like I did but I don't know if the reasoning is right. In other words, in this answer https://math.stackexchange.com/a/4381310/870351 why is there a factor $\binom{2n}{2k}$ that I missed ?

Kilkik
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    If you have $2n$ steps and you choose $k$ of them to be N, that leaves you with $2n - k$ steps to be divided among S, E, W. (Also your description of how your answer differs from the linked correct answer is not right.) – JBL May 17 '23 at 00:28
  • @JBL Yes perfect thanks ! – Kilkik May 17 '23 at 00:36

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Thanks to @JBL's comment, I understood what was wrong. I tried to copy too much the argument in $\Bbb Z$ and got confused. In $\Bbb Z$, after choosing $n$ steps to the right, then all the remaining $n$ steps are in the left direction. In $\Bbb Z^2$ however, after choosing $k$ steps so to the right, then the $k$ steps to the left still need to be chosen among the $2n-k$ remaining steps and in fact, $$\sum_{k=0}^n \binom{2n}{k}\binom{2n-k}{k}\binom{2n-2k}{n-k} $$ is the right answer.

Kilkik
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